Finding eigenvectors when one column is zero

Question

I have the matrix: $\begin{pmatrix} 3 & a & 0 \\ 0 & 4 & 1 \\ 0 & 2 & 5 \\ \end{pmatrix}$

and I should find every eigenvector for every eigenvalue.

After calculations I get that my eigenvalues are: $\lambda_1 = 6 $ and $ \lambda_2 = 3 = \lambda_3 $.

The eigenvector for $\lambda_1$ is: $t\begin{pmatrix} a\\3\\6 \end{pmatrix}$.

However, for $\lambda = 3$ I get (after row reduction): $\begin{pmatrix} 0 & a & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \\ \end{pmatrix}$ which gives me:

$ax_2 = 0$ and $x_2 = - x_3$

The problem is that I don't really know where to go from here. I mainly have 2 questions:

1. How do I handle the variable a?

2. What do I do with $x_1 = 0$ ? (Especially unsure about this in general when it comes to eigenvectors)

Answer 1

Assuming your prior calculations are correct, the situation hinges on whether $a$ is zero or not. If $a \neq 0$, then $x_2$ is forced to be zero, so $x_3$ is as well, and $x_1$ is free. If $a=0$, then one of $x_2$ and $x_3$ can be made free, while $x_1$ is again free.

Answer 2

If $a \ne 0$ then $\pmatrix{a \\ 1 \\ 0}$ and $\pmatrix{0 \\ 1 \\ 0}$ are linearly independent so the eigenspace is given only by $\operatorname{span}\left\{\pmatrix{1 \\ 0 \\ 0}\right\}$, since the rank of your matrix is $2$.

If $a = 0$ then the second and third columns are equal so the eigenspace is $\operatorname{span}\left\{\pmatrix{1 \\ 0 \\ 0}, \pmatrix{0 \\ 1 \\ -1}\right\}$, since rank of your matrix is now $1$.

Answer 3

Solving

$$\begin{pmatrix} 0 & a & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \\ \end{pmatrix}v=0$$

for $a=0$ we obtain $(0,-1,1)$ and $(1,0,0)$ while for $a \neq 0$ we obtain $(1,0,0)$.